Inductive definition of group cohomology?

Question

At the start of Atiyah and Wall's section on group cohomology (in the Cassels-Frhlich collection of Algebraic Number Theory notes) they, of course, define group cohomology (actually, a 'cohomological extension' of the functor $A^G$ which takes the form of cohomology groups) in terms of free resolutions of the $G$-module $\Bbb Z$. To prove that this extension is unique up to canonical equivalence/that the cohomology groups don't depend on the resolution taken, the authors shows that there's an equivalent inductive definition of the cohomology groups, as follows:

Fix a $G$-module $A$, and consider the $G$-module $A^* := \text{Hom}(\Bbb Z[G], A)$, and $A'=A^*/A$. We have an exact sequence $0 \rightarrow A \rightarrow A^* \rightarrow A' \rightarrow 0$. Recalling the definition of a cohomological extension, we know that $H^q(A^*,G)$ vanishes for positive $q$ since $A^*$ is co-induced; and since our short exact sequence induces a long exact sequence of cohomology groups, we have that $$\delta : H^q(G, A') \rightarrow H^{q+1}(G,A)$$ is an isomorphism for positive $q$, and that $$H^1(G,A) \cong \text{Coker}(H^0(G,A^*) \rightarrow H^0(G,A'))$$ Of course, by definition $H^0(G,A) = A^G$. This is all good so far. But now Atiyah and Wall lose me: they say that all of the $H^q$ can be constructed inductively from $H^0$. I don't see how this is the case, at least with what I have here. I would need some way to pass from $H^q(G,A)$ to $H^q(G,A')$ (EDIT: not just homomorphically, which is easy, but in some way that completely determines the latter group) to actually know ALL of the cohomology groups, but I don't see a mechanism by which I can do this. What am I missing here?

If I need to further explain the terminology, let me know (I tried to keep it to a minimum lest I simply reproduce the first two pages).

Answer 1

The inductive argument is:

if we know $H^i(G, A)$ for all $A$, then we know $H^{i+1}(G, A)$ for all $A$. In other words, the inductive hypothesis is much stronger than saying that we know $H^i(G, A)$ for one particular $A$.

Answer 2

You said it actually : because $\delta:H^q(G,A')\rightarrow H^{q+1}(G,A)$ is an isomorphism you can get all the cohomology group. It works as follows.

Assume you can construct $H^q(G,A)$ for all abelian group $A$, then you can construct $H^{q+1}(G,A)$ for all abelian group $A$.

Indeed, you can construct $H^0(G,A)$ for all abelian group $A$ since it is just $A^G$. To construct $H^1(G,A)$, just take $\mathrm{Coker}(H^0(G,A^*)\rightarrow H^0(G,A')$. You thus have $H^1(G,A)$ for all abelian group $A$.

Assume you can construct $H^q(G,A)$ for all abelian group $A$. Take any $A$, and construct $A^*,A'$ as before. By induction you can construct $H^q(G,A')$ and because $\delta:H^q(G,A')\rightarrow H^{q+1}(G,A)$ is an isomorphism, you have $H^{q+1}(G,A)$ for $A$.

Note that $A\mapsto A^*, A'$ are functorial in $A$ so that $H^q(G,A)$ are really functors of $A$.